Is $X^2$ independent from $XY$ where $X$ and $Y$ are standard normals?

Question

I'm thinking they can somehow be expressed as functions of $X-Y$ and $X+Y$, but I haven't quite found out how.

Bonus questions:

• Is it correct that they are both Chi square distributed?

• And so, would $X^2 + XY$ also be Chi square distributed?

• What would be the distribution of $a(X_0^2+Y_0^2) + b(X_0X_1+Y_0Y_1)$?

Answer 1

Let $X$ and $Y$ be independently $Norm(0, 1),$ and define $Q = X^2$ and $T = XY.$

The following simulation suggests that $Cor(Q, T) = 0,$ and clearly shows that $Q$ and $T$ are not independent.

 m = 10^5;  x = rnorm(m);  y = rnorm(m)
 q = x^2;  t = x*y
 cor(q,t)
 ## 0.00575081  # consistent with 0 correlation
 plot(q, t, pch=".")
 abline(h = 5, col="green2", lwd=2)
 abline(v = 2, col="green2", lwd=2)

The figure seems to show that $P(Q < 1) > 0,\,$ and $P(T > 7) > 0,\,$ but $P(Q < 1, T > 7) \approx 0.$ I guess something like this may be simple to prove rigorously.

Also, it seems clear that $T$ is not chi-squared because it takes both positive and negative values.

You also have some useful hints in the comments.